Systems with three variables
Systems of Three Variables 2 Systems of Three Variables 2
⇐ Use this menu to view and help create subtitles for this video in many different languages.
You'll probably want to hide YouTube's captions if using these subtitles.
- Solve this system. And once again, we have three equations
- with three unknowns, so this is essentially trying to find out where
- three different planes would intersect in three dimensions.
- And to do this, if you want to do by elimination, if we want to eliminate variables
- it looks like -- well, it looks like we have a negative z here we have
- a plus 2z we have a 5z over here.
- If we were to scale up this third equation by positive 2,
- then you'd have a negative 2z here and would cancel out
- with this 2z there, and if you were to scale it up by 5, you'd
- have a negative 5z here and that can cancel out with that 5z over there.
- So let's try to cancel out; let's try to eliminate the z's first.
- So let me start with this equation up here;
- I'll just rewrite it, so we have -- I'll draw an arrow over here --
- we have x + 2y + 5z is equal to -17, and then to cancel out or to eliminate the z's,
- I'll multiply this equation here times 5.
- So I'm going to multiply this equation times 5. So we have to multiply both sides by 5.
- So 3x times 5 is 15x, y times 5 is plus 5y, and then -z times 5 is -5z --
- that's the whole point why we're multiplying by 5 -- is equal to 3 times 5
- which is equal to 15. And so if we add these two equations,
- we get x plus 15x is 16x, 2y plus 5y is 7y, and 5z minus 5z or plus negative 5z --
- those are going to cancel out -- and that is going to be equal to
- negative 17 plus 15 is negative 2.
- So we're able to use the constraints in that equation and that equation
- and now we have the equation with just x and y.
- So let's try to do the same thing -- let's try to eliminate the z's --
- but now use this equation and this equation.
- So this equation -- let me just rewrite it over here --
- we have 2x minus 3y plus 2z is equal to negative 16 and I just rewrote it.
- And now, so that this 2z gets eliminated, let's multiply this equation times 2.
- So let's multiply it times 2 so we'll have a negative 2z here to eliminate with the positive 2z.
- So 2 times 3x is 6x, 2 times y is plus 2y, and 2 times negative z is negative 2z
- is equal to 2 times 3 is equal to 6. And now we can add these two equations.
- 2x plus 6x is 8x, negative 3y plus 2y is negative y, and then these two guys get cancel out,
- and then that is equal to negative 16 plus 6 is negative -- is negative 10.
- So now we have two equations with two unknowns; we eliminated the z's.
- And, let's see, if we want to eliminate again,
- we have a negative y over here, we have a positive 7y, we could have eliminated the y's
- if we multiply this times 7, and add the two equations. So let's do that.
- So let's multiply this times 7. 7 times 8 is 56, so it's 56x minus 7y is equal to
- 7 times negative 10 is equal to negative 70, and now we can add these two equations.
- I'm now trying to eliminate the y's. So we have 16x plus 56x, that is 72x.
- So we have 72x, these guys eliminate, equal to negative 72.
- Negative 2 plus negative 70. Divide both sides by 72, and we get x is equal to negative 1.
- And now we just have to substitute back to figure out what y and z are equal to.
- So let's go back to this equation right over here,
- we have 8x minus y is equal to negative 10.
- If x is equal to negative 1, that means 8 times negative 1 or negative 8
- minus y is equal to negative 10. We can add 8 to both sides, and so we have
- negative y is equal to negative 2. Or multiplying both sides by negative 1,
- y is equal to 2. Let me square that off. So x is equal to negative 1, y is equal to 2,
- we now just have to worry about z, and we can go back to any of these up here.
- So let's just use -- I'll just use the last numbers just because they seem lower --
- so if we substitute back into the last equation right over here,
- we have 3 times x, which is 3 times negative 1, plus y, which is 2, minus z is equal to 3.
- So negative 3 plus 2 minus z is equal to 3.
- And this is negative 1 minus z is equal to 3, and then add one to both sides,
- we get -- these cancel out -- negative z is equal to 4, multiplying both sides by negative 1,
- you get z is equal to negative 4.
- So we're done. Let's verify that these solutions or this solution of
- x is negative 1, y is equal to 2, z is equal to negative 4 actually satisfies
- all three of these constraints. So let's substitute into the first one.
- So we have x + 2y + 5z. So that is x is negative 1 plus 2 times y, so plus 4,
- plus 5z, so minus 20 has to be equal to negative 17.
- And this is negative, this right here is positive 3 minus 20 is indeed equal to negative 17.
- So it satisfies the first constraint. So the second one, 2 times x, 2 times negative 1, that's negative 2,
- minus 3 times y, that's minus 6, plus 2 times z, z is equal to negative 4,
- so that's 2 times negative 4 is negative 8. That needs to be equal to negative 16.
- Negative 2 minus negative 8 is -- sorry -- negative 2 minus negative 6 is negative 8,
- subtract another 8, you get negative 16. So it meets the second constraint.
- And then finally, let's look at the last constraint. We have 3 times x,
- so negative 3 plus y, so plus 2, minus z, so minus negative 4 is the same thing as plus 4,
- that needs to be equal to 3. So negative 3 plus 2 is negative 1, plus 4 is indeed equal to 3.
- So we found our point of intersection in three dimensions, at these 3 planes.
- x is negative 1, z is negative 4, y is 2, and we're able to verify it that it does
- indeed meet all of the constraints.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
|
Have something that's not a question about this content? |
This discussion area is not meant for answering homework questions.
Discuss the site
For general discussions about Khan Academy, visit our Reddit discussion page.
Flag inappropriate posts
Here are posts to avoid making. If you do encounter them, flag them for attention from our Guardians.
abuse
- disrespectful or offensive
- an advertisement
not helpful
- low quality
- not about the video topic
- soliciting votes or seeking badges
- a homework question
- a duplicate answer
- repeatedly making the same post
wrong category
- a tip or feedback in Questions
- a question in Tips & Feedback
- an answer that should be its own question
about the site
Share a tip
Suggest a fix
Have something that's not a tip or feedback about this content?
This discussion area is not meant for answering homework questions.